Problem: Simplify and expand the following expression: $ \dfrac{2}{2q + 14}+ \dfrac{1}{q + 2}- \dfrac{3}{q^2 + 9q + 14} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{2}{2q + 14} = \dfrac{2}{2(q + 7)}$ We can factor the quadratic in the third term: $ \dfrac{3}{q^2 + 9q + 14} = \dfrac{3}{(q + 7)(q + 2)}$ Now we have: $ \dfrac{2}{2(q + 7)}+ \dfrac{1}{q + 2}- \dfrac{3}{(q + 7)(q + 2)} $ The least common multiple of the denominators is: $ 2(q + 7)(q + 2)$ In order to get the first term over $2(q + 7)(q + 2)$ , multiply by $\dfrac{q + 2}{q + 2}$ $ \dfrac{2}{2(q + 7)} \times \dfrac{q + 2}{q + 2} = \dfrac{2(q + 2)}{2(q + 7)(q + 2)} $ In order to get the second term over $2(q + 7)(q + 2)$ , multiply by $\dfrac{2(q + 7)}{2(q + 7)}$ $ \dfrac{1}{q + 2} \times \dfrac{2(q + 7)}{2(q + 7)} = \dfrac{2(q + 7)}{2(q + 7)(q + 2)} $ In order to get the third term over $2(q + 7)(q + 2)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{3}{(q + 7)(q + 2)} \times \dfrac{2}{2} = \dfrac{6}{2(q + 7)(q + 2)} $ Now we have: $ \dfrac{2(q + 2)}{2(q + 7)(q + 2)} + \dfrac{2(q + 7)}{2(q + 7)(q + 2)} - \dfrac{6}{2(q + 7)(q + 2)} $ $ = \dfrac{ 2(q + 2) + 2(q + 7) - 6} {2(q + 7)(q + 2)} $ Expand: $ = \dfrac{2q + 4 + 2q + 14 - 6}{2q^2 + 18q + 28} $ $ = \dfrac{4q + 12}{2q^2 + 18q + 28}$ Simplify: $ = \dfrac{2q + 6}{q^2 + 9q + 14}$